Period of 3sin(7πn/5 + π/2)

$x[n] = 3\sin\left(\frac{7\pi}{5}n + \frac{\pi}{2}\right)$, with $\omega_0 = \frac{7\pi}{5}$.

Check periodicity:

$$\frac{\omega_0}{2\pi} = \frac{7\pi/5}{2\pi} = \frac{7}{10} \quad\text{(rational, so periodic)}$$

Fundamental period:

$$N = \frac{2\pi m}{\omega_0} = \frac{2\pi m}{7\pi/5} = \frac{10m}{7}$$

The smallest integer $N$ is obtained with $m = 7$:

$$N = 10 \text{ samples}$$

Fundamental frequency:

$$\omega = \frac{2\pi}{N} = \frac{2\pi}{10} = \frac{\pi}{5}\ \text{rad/sample}, \qquad f = \frac{1}{N} = \frac{1}{10} = 0.1\ \text{cycles/sample}$$

Check: $x[n+10] = 3\sin\left(\frac{7\pi}{5}n + 14\pi + \frac{\pi}{2}\right) = x[n]$ since $14\pi$ is a multiple of $2\pi$. The phase $\pi/2$ does not affect the period.

Answer: $N = 10$, fundamental frequency $\pi/5$ rad/sample (0.1 cycles/sample).